What is the expected number of times one must roll a fair dice before getting \(2\) consecutive \(6\)’s?
According to the geometric distribution, we can expect \(6\) rolls before landing the first \(6\). At that point, there is a \(\frac{1}{6}\) chance that the next roll is a \(6\), which would result in two consecutive \(6\)’s. On the other hand, there is a \(\frac{5}{6}\) chance of landing a number that is not \(6\), at which point we must reset our count. This formulation can be represented by the following recurrence, where \(X\) is a random variable representing the number of rolls for \(2\) consecutive \(6\)’s:
\[E[X] = 6 + \frac{1}{6}(1) + \frac{5}{6}(1+E[X])\] \[E[X] = 42\]
Thus, the expected number of rolls to land \(2\) consecutive \(6\)’s is \(42\).
This approach can be generalized to an arbitrary number of consecutive \(6\)’s. For example, suppose \(Y\) represents the number of rolls for \(3\) consecutive \(6\)’s. Knowing that \(E[X] = 42\), we can write a recurrence for \(E[Y]\) as follows:
\[E[Y] = E[X] + \frac{1}{6}(1) + \frac{5}{6}(1+E[Y])\] \[E[Y] = 258\]